Compact and countability axioms!

Question

I wondering which countability axioms compact imply in arbitrary topological spaces. I'm using Greene/Gamelin 2nd ed. And they list separable, 2nd-coutable, first-countable and Lindelöf. Clearly compact implies Lindelöf but does it imply any of the others?

Answer 1

Compact does not imply any of the others. Consider the extended long line, which is compact but not separable or first-countable (hence not second-countable). It is compact by definition (it is the two-point compactification of the long line). To see that it is not separable, note that for any countable collection of points $x_n$ we have some $\lambda\in \omega_1$ such that $x_n\notin\{\lambda\}\times (0,1)$ for all $n$ by the uncountability of $\omega_1$, and $\{\lambda\}\times (0,1)$ is open so the points $x_n$ are not dense. To see that it is not first-countable, let $U_n$ be a countable neighborhood basis for $+\infty$. Since each $U_n$ is open and contains $+\infty$, we have some $\lambda_n\in\omega_1$ such that $(\lambda_n,+\infty]\subseteq U_n$. But since $\omega_1$ is uncountable, we have some $\lambda\in \omega_1$ such that $\lambda>\lambda_n$ for all $n$, thus $U_n\not\subseteq (\lambda, +\infty]$, thus the $U_n$ are not a countable neighborhood basis for $+\infty$.

Answer 2

For an even simpler example let $D$ be any uncountable set, $p$ a point not in $D$, and $X=\{p\}\cup D$, and take

$$\mathscr{B}=\big\{\{x\}:x\in D\big\}\cup\{X\setminus F:F\subseteq D\text{ is finite}\}$$

as a base for the topology on $X$. ($X$ is the one-point compactification of $D$, where $D$ has the discrete topology.)

• $X$ is compact, because every open set containing $p$ has a finite complement in $X$.
• $X$ is not first countable, because $p$ does not have a countable local base. Let $\mathscr{U}=\{U_n:n\in\Bbb N\}$ be a countable family of open nbhds of $p$. For $n\in\Bbb N$ let $F_n=X\setminus U_n$, and let $H=\bigcup_{n\in\Bbb N}F_n$; each $F_n$ is finite, so $H$ is countable, and $D\setminus H\ne\varnothing$. Fix $x\in D\setminus H$, and let $V=X\setminus\{x\}$; then $V$ is an open nbhd of $p$, but $x\in U_n\setminus V$ for each $n\in\Bbb N$, so $V$ contains no member of $\mathscr{U}$, and $\mathscr{U}$ therefore is not a local base at $p$.
• Every second countable space is trivially first countable, so $X$ is not second countable.